Problem: ${\sqrt[3]{80} = \text{?}}$
$\sqrt[3]{80}$ is the number that, when multiplied by itself three times, equals $80$ First break down $80$ into its prime factorization and look for factors that appear three times. So the prime factorization of $80$ is $2\times 2\times 2\times 2\times 5$ Notice that we can rearrange the factors like so: $80 = 2 \times 2 \times 2 \times 2 \times 5 = (2\times 2\times 2) \times 2\times 5$ So $\sqrt[3]{80} = \sqrt[3]{2\times 2\times 2} \times \sqrt[3]{2\times 5}$ $\sqrt[3]{80} = 2 \times \sqrt[3]{2\times 5}$ $\sqrt[3]{80} = 2 \sqrt[3]{10}$